WHY PLANETS HAVE ELLIPTICAL ORBITS

1. WHY PLANETS HAVE ELLIPTICAL ORBITS REYNALDO B. VEA PHILIPPINE SCIENCE HIGH SCHOOL NOVEMBER 10, 2008

2. QUIZ THE EARTH IS CLOSEST TO THE SUN IN THE MONTH OF JANUARY THE EARTH IS FARTHEST FROM THE SUN IN THE MONTH OF JULY

3. ADDITION OF VECTORS A + B = C B C C B A A TIP-TO-TAIL ADDITION PARALLELOGRAM LAW

4. VELOCITY IS A VECTOR QUANTITY BUT HERE RELATIVE TO GROUND DUE TO CURRENT NOT HERE RELATIVE TO WATER CURRENT

5. FORCE IS A VECTOR QUANTITY 2 MULES PULLING A BOAT ON EITHER BANK

6. FORCE IS A VECTOR QUANTITY SAME AS ONE RIVER TUG PULLING THE BOAT STRAIGHT UP THE MIDDLE

7. TYCHO BRAHE URANIBORG MURAL QUADRANT SEXTANT

8. Kepler’s Laws: • Each planet moves around the sun in an ellipse with the sun at one focus. • The radius vector from the sun to the planet sweeps out equal areas in equal intervals of time. • The squares of the period of any two planets are proportional to the cubes of the semimajor axes of their respective orbits: T a a3/2.

9. IN ONE OF HIS EXPERIMENTS GALILEO FOUND THAT IF A BALL WAS ALLOWED TO ROLL DOWN ONE PLANE AND BACK UP ANOTHER IT WOULD TEND TO KEEP ROLLING UP THE SECOND PLANE UNTIL IT REACHED THE SAME HEIGHT IT STARTED AT. WITH A LEAP OF THE IMAGINATION, AGAIN, HE STATED THE IF THE SECOND PLANE WERE HORIZONTAL THE BALL WOULD NEVER STOP ROLLING (BECAUSE IT WOULD NEVER REACH THE ORIGINAL HEIGHT). THUS HE CONCLUDED THAT THE NATURAL STATE OF AN OBJECT IN HORIZONTAL MOTION WAS TO KEEP ON MOVING HORIZONTALLY, AT CONSTANT SPEED, FOREVER. THIS LATER WOULD LEAD TO THE LAW OF INERTIA. THIS IS A RADICAL DEPARTURE FROM ARISTOTELIAN PHILOSOPHY IN WHICH ANY HORIZONTAL MOTION REQUIRED A PROXIMATE CAUSE.

10. RENE DESCARTES GENERALIZES GALILEO’S LAW OF INERTIA TO NOT ONLY APPLY TO HORIZONTAL MOTION THUS: IN THE ABSENCE OF ANY EXTERNAL FORCE ACTING ON IT A BODY AT REST WILL REMAIN AT REST AND A BODY IN MOTION WILL REMAIN IN MOTION, AT CONSTANT SPEED, IN A STRAIGHT LINE.

11. NEWTON’LAWS OF MOTION LAW I. EVERY BODY CONTINUES IN ITS STATE OF REST, OR OF UNIFORM MOTION IN A STRAIGHT LINE, UNLESS IT IS COMPELLED TO CHANGE THAT STATE BY FORCES IMPRESSED UPON IT. LAW II. THE CHANGE IN MOTION IS PROPORTIONAL TO THE MOTIVE FORCE IMPRESSED; AND IS MADE IN THE DIRECTION OF THE STRAIGHT LINE IN WHICH THAT FORCE IS IMPRSSED. LAW III. TO EVERY ACTION THERE IS ALWAYS OPPOSED AN EQUAL REACTION; OR, THE MUTUAL ACTION OFBTWO BODIES UPON EACH OTHER ARE ALWAYS EQUAL, AND DIRECTED TO CONTRARY PARTS.

12. PREDICTION Kepler’s 1st Law: The Law of Ellipses Kepler’s 2nd Law: The Law of Areas All forces are directed towards the sun Force of gravity Forces are inversely proportional to the square of the distances Kepler’s 3rd Law: T a a 3/2 Descartes law of Inertia Newton’s 1st Law: The Law of Inertia Newton’s 2nd Law: F = d(mv)/dt Galileo’s law of inertia Newton’s 3rd Law: Action and reaction

13. P F’ F

14. Q P q F’ F

15. F’ F ALL LIGHT RAYS STARTING AT ONE FOCUS WILL BE FOCUSED TO A POINT AT THE OTHER FOCUS.

16. TANGENT LINE P ELLIPSE OR ANY CURVE

17. TANGENT LINE P reflected light ray light ray ELLIPSE OR ANY CURVE LIGHT RAY MIRROR-REFLECTED FROM THE CURVE

18. TANGENT LINE P reflected light ray light ray ELLIPSE OR ANY CURVE LIGHT RAY MIRROR-REFLECTED FROM THE TANGENT LINE

19. A LIGHT RAY, MIRROR-REFLECTED FROM FROM THE CURVE AT ANY POINT, FOLLOWS THE SAME PATH IF IT WERE MIRROR-REFLECTED AT THAT POINT FROM THE TANGENT LINE. TANGENT LINE P reflected light ray light ray THE REASON THAT LIGHT REFLECTS FROM THE CURVE JUST AS IT WOULD FROM THE TANGENT LINE AT THE SAME POINT IS THAT THE TANGENT LINE INDICATES THE DIRECTION OF THE CURVE AT EXACTLY THAT POINT. ELLIPSE OR ANY CURVE

20. THIS ANGLE MIRROR EQUALS THIS ANGLE REFLECTED LIGHT RAY LIGHT RAY LAW OF REFLECTION FROM A FLAT MIRROR

21. THIS ANGLE EQUALS THIS ANGLE P REFLECTED RAY INCIDENT RAY F’ F THE INCIDENT RAY FROM F TO P MAKES THE SAME ANGLE WITH THE TANGENT LINE AT P AS DOES THE REFLECTED RAY, WHICH GOES TO F’. PROVE THAT THIS STATEMENT IS EQUIVALENT TO SAYING THAT THE DISTANCE F’P PLUS THE DISTANCE FP IS THE SAME FOR ANY POINT P ON THE ELLIPSE.

22. START OF PROOF G’ T P F’ F

23. G’ T P F’ G’T = F’T TP = TP ANGLE GTP=ANGLE F’TP = RT ANGLE SAS CONGRUENCE

24. G’ T P F’ G’P = F’P CONGRUENCE ANGLE F’PT = ANGLE G’PT

25. ANGLE F’PT = ANGLE G’PT ANGLE G’PT = ANGLE FPS G’ therefore ANGLE F’PT = ANGLE FPS perpendicular bisector perpendicular bisector line ST will reflect light form F to F’ at point P P T S F’ F F’P = G’P CONGRUENCE therefore F’P + PF = FG’

26. G’ straight string F’ F

27. F’ F same string bent

28. G’ perpendicular bisector P F’ F

29. G2’ G3’ G1’ G4’ F’ F

30. WE HAVE A LINE THAT REFLECTS LIGHT THAT ARRIVES FROM POINT F AT POINT P WITH EQUAL ANGLES OF INCIDENCE AND REFLECTION BACK TO POINT F’. THIS LINE HAPPENS TO BE THE PERPENDICULAR BISECTOR OF F’G’. G’ perpendicular bisector (reflecting line) P T F’ F WE ALSO HAVE AN ELLIPSE THAT OBEYS THE STRING-AND-TACKS CONSTRUCTION, i.e., F’P + PF IS THE SAME ALL THE WAY AROUND THE STRING. RECAPITULATION

31. WE KNOW THAT EACH POINT ON THE ELLIPSE HAS THE SAME MIRROR-REFLECTING PROPERTIES AS A TANGENT LINE AT THAT POINT. G’ perpendicular bisector (reflecting line) Also tangent? THUS IF THERELECTING LINE AT P IS ALSO TANGENT TO THE ELLIPSE AT P THEN WE HAVE PROVED THAT THE TWO PROPERTIES (STRING-AND-TACKS CONSTRUCTION AND REFLECTING LIGHT FROM ONE FOCUS TO THE OTHER) ARE EQUIVALENT. P T F’ F ALL THAT’S LEFT TO PROVE, THEREFORE, IS THAT THE REFLECTING LINE AT POINT P IS ALSO TANGENT TO THE ELLIPSE AT POINT P.

32. TANGENT: EVERY POINT BUT P IS OUTSIDE THE ELLIPSE P F’ F NONTANGENT: THIS SEGMENT IS INSIDE THE ELLIPSE

33. F’Q = G’Q G’ F’Q +QF = G’Q + QF Q P T F’ F

34. G’Q + QF > G’P + PF G’ since G’PF is a straight line Q P F’ F

35. G’Q + QF > G’P + PF earlier it was shown that: F’Q +QF = G’Q + QF F’P + PF = G’P +FP Q therefore: F’QF = G’QF > G’PF = F’PF P or F’QF > F’PF for any such point q on that line except for the one coincident with point P F’ F IN OTHER WORDS IF WE WANTED TO REACH POINT Q WITH A STRING STRETCHED FROM TACKS AT F AND F’, THE STRING WOULD HAVE TO BE LONGER THAN THE ONE NEEDED TO REACH THE UNIQUE POINT P. IT WAS SHOWN EARLIER THATTHIS MEANS ALL SUCH POINTS ARE OUTSIDE THE ELLIPSE. THUS THE LINE IS TANGENT TO THE ELLIPSE AT POINT P. QED !

36. PART II B S A DURING A CERTAIN INTERVAL OF TIME

37. c Bc = AB B S A DURING NEXT INTERVAL OF TIME with no force acting

38. Sun’s force represented by an impulse applied at B results in a component of motion directed toward the Sun, BV compounding of Bc and BV using parallelogram law yields “actual” motion BC note: Cc is not directed toward the Sun but is parallel to BV C c V B S A DURING NEXT INTERVALS OF TIME with Sun’s force acting

39. Sun’s force represented by an impulse applied at B results in a component of motion directed toward the Sun, BV compounding of Bc and BV using parallelogram law yields “actual” motion BC note: Cc is not directed toward the Sun but is parallel to BV same procedure is repeated at each point D d C c V B S A DURING NEXT INTERVALS OF TIME with Sun’s force acting

40. SHOW THAT: area of D SAB = area of D SBC C c B S A SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

41. SHOW THAT: area of D SAB = area of D SBC First step is to show that area of D SAB = area of D SBc C c B S A SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

42. SHOW THAT: area of D SAB = area of D SBC First step is to show that area of D SAB = area of D SBc AB = Bc ; rt angles equal ; angle ABx = angle cBy congruence of 2 small D’s, by ASA Ax = cy or equal altitudes area of D SAB = area of D SBc QED ! c common base x y altitudes B S A SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

43. SHOW THAT: area of D SAB = area of D SBC First step is to show that area of D SAB = area of D SBc Next show that area of D SBc = area of D SBC Cc parallel to SB altitudes are equal common base SB & equal altitudes area of D SBc = area of D SBC by transmissivity, area of D SAB = area of D SBC C c QED ! B S A SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

44. The path Bc would have been taken if there were no force at all. Instead, there is a force, directed toward S. That force changes the direction from path Bc to path BC, but it cannot change the area swept out during the fixed interval of time. The same idea could be applied to successive triangles swept out in equal intervals of time. We have thus succeeded in proving Kepler’s Second Law of planetary motion! D C c B S A SWEEPING OUT EQUAL AREAS IN EQUAL INTERVALS OF TIME

45. We have thus far used Newton’s first Law (the law of inertia), Newton’s second law (any change in motion is in the direction of the impressed force), and the idea that the force of gravity on the planet is directed toward the Sun. Nothing else. For example, we have not used the idea that the force of gravity is inversely proportional to the square of the distance. So the inverse-square-of-the-distance character of gravity has nothing to do with Kepler’s second law. Any other kind of force would have produced the same result provided only that the force is directed toward the sun. What we have learned is that if Newton’s first and second laws are correct, then Kepler’s observation that planets sweep out equal areas in equal intervals of time means that the gravitational force of the planet is directed towards the sun. Incidentally, we have also used Newton’s first corollary to his laws – that the net motion produced by both tendencies in the time interval is given by the parallelogram of the separate motions that would have occurred. The above demonstration is an exact copy of the one in Principia Mathematica by Newton.

46. NEWTON’S DRAWING

47. FEYNMAN’S DIAGRAM

48. Let T = time for planet to complete one orbit T1/T2 = (a1/a2)3/2 planet 2 planet 1 a1 sun a2

49. b a F’ F

50. A circle is a special case of an ellipse that can be constructed by moving the two foci to the center. Kepler’s laws allow planetary orbits to be circles but don’t require it. In reality the orbits of the planets in our solar system are very nearly circles. R R T a R3/2